= Solution
A <closed curve> has $\gamma(a)=\gamma(b)$. A <simple curve> has no repeated points apart from the permitted equality of its two endpoints when it is closed. A <smooth curve>, in the regular contour-integration convention, has a continuous nonzero tangent $\gamma'$; piecewise smooth contours are treated by adding the integrals over their smooth pieces. For a continuous integrand, its <contour integral> is defined by
$$
\boxed{\int_\gamma f(z)\,dz=\int_a^b f(\gamma(t))\gamma'(t)\,dt.}
$$
For the positively oriented circle, $\gamma'(t)=ire^{it}$ and $(\gamma(t)-z_0)^n=r^ne^{int}$. Direct parameter integration therefore gives
$$
\int_\gamma(z-z_0)^n\,dz=ir^{n+1}\int_0^{2\pi}e^{i(n+1)t}\,dt
=\boxed{\begin{cases}2\pi i,&n=-1,\\0,&n\ne-1.\end{cases}}
$$
For $n\ne-1$, an elementary <antiderivative> is $e^{i(n+1)t}/[i(n+1)]$ and its endpoint values agree because $n$ is an integer. For $n=-1$, the integrand of the parameter integral is the constant $i$. No integral theorem is needed.
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